1 g produs s-a obtinut la 50% dar la 100%
50% ................ 1 g
100% .............. m = 2 g
V L 2 g
CH2=CH2 + [O] + H2O --> HO-CH2-CH2-OH /x3
22,4x3 62x3
n moli
2KMnO4 + H2O --> 2KOH + 2MnO2 + 3[O]
2
V = 22,4x3x2/62x3 = 0,722 L etena in c.n. la 100%
n = 2x2/62x3 = 0,0215 moli KMnO4
Cm = n/Vs => Vs = n/Cm = 0,0215/1 = 0,0215 L = 21,50 mL